# Find the derivative of the function.

1. Feb 22, 2005

Can you work this out step by step so I can see how to do it. Thank you
h(t) = (t^4 - 1)^3(t^3+1)^4

Last edited: Feb 22, 2005
2. Feb 22, 2005

### saltydog

It's a bit ambiguous. Do you mean:

$$(t^4-1)^3 (t^3+1)^4$$

I think so. In that case, need to use the Chain Rule. If it confussing to you, try a simpler problem first (I do that too). For example, try this one:

$$(t^2-1)t^3$$

Wouldn't that just be:

$$(t^2-1)3t^2+t^3(2t)$$

3. Feb 23, 2005

### CrankFan

Well you're going to use the product rule, so first use the chain rule to get the derivative of each factor:

$$\left[ (t^4 - 1)^3 \right]^{\prime} = 3(t^4 - 1)^2 4t^3 = 12t^3(t^4-1)^2$$

$$\left[ (t^3 + 1)^4 \right]^{\prime} = 4(t^3 + 1) 3t^2 = 12t^2(t^3 + 1)^3$$

Then apply the product rule:

$$12t^3(t^4-1)^2 (t^3 + 1)^4 + 12t^2(t^3 + 1)^3 (t^4 - 1)^3$$

4. Feb 23, 2005

### Tom Mattson

Staff Emeritus
We don't do that here. At Physics Forums you have to show your work to receive homework help.